## Integration

Here is the Problem:

$$\int_0^\pi\theta^2cosec^2\theta d\theta$$

I want to do this problem by parts and finding hard to find shortest way to do the problem any shortest way pls
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 Recognitions: Science Advisor The straightforward way is probably the easiest. integral of csc2 is -ctn and integral of ctn is log(sin). Integrating log(sin) may take a little work.

 Originally posted by himanshu121 Here is the Problem: $$\int_0^\pi\theta^2cosec^2\theta d\theta$$ I want to do this problem by parts and finding hard to find shortest way to do the problem any shortest way pls
cos*sec^2(x) or cosx*sec^2(x)?

## Integration

 Originally posted by PrudensOptimus cos*sec^2(x) or cosx*sec^2(x)?
Isn't it csc(x)?

It is cosecant(x) i.e csc(x)
 Integrating log(sin) may take a little work
This is probably the easiest way of doing which i too have tried but it is not the shortest way

There are many ways of doing a problem i am looking for shortest way
Thnxs
 How about: \begin{align*} \int_0^\pi\theta^2\csc^2\theta\;d\theta &=\int_0^1\theta^2\csc^2\theta\;d\theta+\int_1^\pi\theta^2\csc^2\theta\ ;d\theta \\ &>\int_0^1\theta^2\csc^2\theta\;d\theta+\int_1^\pi\csc^2\theta\;d\th eta \\ &=\int_0^1\theta^2\csc^2\theta\;d\theta+\left[-\cot\theta\right]_1^\pi \end{align*} Now since $\lim\limits_{\theta\rightarrow\pi}(-\cot\theta)=+\infty$ we know that the rightmost term diverges. Moreover, the left integral (the one from 0 to 1) is clearly positive. Thus the original integral clearly diverges.
 master_coda: the integral of x^2*(cscx)^2 from 1 to pi comes out to be about 4.2 the integral of (cscx)^2 from 1 to pi goes to inifinite. I believe to say the integral of x^2*(cscx)^2 from 1 to pi is greater than the integral of (cscx)^2 from 1 to pi since according to the values it came out as then the formula should be switched around and divergence would not be proved. himanshu121: Best way to do it would be to plug it into a calculator or look it up in a table. Otherwise you're probably stuck with integrating log(sin(x)). jk

 Best way to do it would be to plug it into a calculator or look it up in a table
We are not allowed to use calculator in India till we are undergraduate
I found the way but dont know whether it is shortest one or not but definetly i wont stuck at log(sinx)

$$I = \int_0^\pi\theta^2\csc^2\theta d\theta$$

$$I= \int_0^\pi\ (\pi-\theta)^2\csc^2\theta d\theta$$

this gives

$$\pi\int_0^\pi\csc^2 \theta d\theta = 2\int_0^\pi\theta\csc^2\theta d\theta$$

integrating by parts with one part as $$\theta$$ and other as $$\theta\csc^2\theta d\theta$$

i will get$$\int_0^\pi \cot\theta d\theta = \log(csc\theta-cot\theta)$$

much easier than integrating log(sinx)

But another problem is how i will put the limits in cotx from 0 to pi in both cases it is infinity and i know there is no break in the function cotx b/w these points

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I agree with master_coda...

 the integral of x^2*(cscx)^2 from 1 to pi comes out to be about 4.2
Where did you come up with this?! My calculator says 3*10^14 with the caveat of "questionable accuracy" (though, 3*10^14 is a good approximation of infinity. [;)])

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 I'm pretty sure that the integral diverges. My algebra, my calculator and my computer all agree. Also, if it's any help, $\csc\theta-\cot\theta=\frac{1-\cos\theta}{\sin\theta}$.